A Jump in the Rate Is Not a Jump in the Accumulation
A rate describes how quickly a quantity changes. If the rate changes abruptly at a particular input, the accumulated quantity does not instantly change by a matching amount. Its change is determined by integrating the rate over an interval, and an interval of vanishing length contributes a vanishing amount when the rate stays bounded.
This distinction is useful whenever a rate is given by different formulas on different intervals. The formulas may meet with a jump at a breakpoint, but the accumulated total must still be calculated from the signed area on each side. We will show why the accumulation stays continuous at the breakpoint and work through examples that cross one or more jumps.
Rates with Finitely Many Jumps
Suppose a rate \(f\) is defined on an interval containing a fixed basepoint \(a\). We call \(f\) piecewise continuous with finitely many breakpoints on a compact interval if that interval can be divided at finitely many points so that \(f\) is continuous on the interior of each resulting piece and has finite one-sided limits at the division points. Values at the breakpoints themselves may be specified separately.
Such a rate is bounded on the compact interval and is integrable there. Its accumulation from \(a\) is the function \(A\) defined by
When \(x<a\), the integral has the usual orientation: \(\int_a^x f(t)\,dt=-\int_x^a f(t)\,dt\). If a breakpoint \(c\) lies between the limits of integration, split the integral there. The contributions from the separate pieces retain their signs.
The value assigned to \(f\) at a single breakpoint does not affect the integral. What matters for the accumulation is the behavior of the rate over intervals, not its value at one isolated point.
Why Accumulation Stays Continuous at a Jump
Let \(c\) be a breakpoint, and suppose \(|f(t)|\leq M\) near \(c\), for some finite constant \(M\). The interval-increment identity gives \(A(x)-A(c)=\int_c^x f(t)\,dt\). The integral bound then shows that the size of this difference is at most \(M\) times the interval length.
This inequality applies on either side of \(c\). In particular, as \(x\) approaches \(c\), the right-hand side approaches zero, so \(A(x)\) approaches \(A(c)\). This is the continuity condition at \(c\). Notice that the argument needs boundedness near the breakpoint; a finite jump satisfies that condition.
The bound gives more than continuity: it controls how quickly the accumulation can change near \(c\). But it does not say that the rate itself is continuous. Nor does continuity of \(A\) guarantee that \(A\) has a derivative at the jump.
To see the distinction, suppose the one-sided limits of \(f\) at \(c\) are \(L_-\) and \(L_+\). For \(h>0\), the difference quotient on the right is
As \(h\) shrinks, this is the average rate on a shorter and shorter interval. If \(f(t)\) approaches \(L_+\) from the right, that average approaches \(L_+\); the corresponding left-hand quotient approaches \(L_-\). Thus a jump in the rate can leave the accumulation continuous while preventing its two-sided derivative from existing when \(L_-\ne L_+\).
Calculating Across a Breakpoint
To calculate an accumulated value across a jump, first find the accumulation up to the breakpoint. Then integrate the formula that applies beyond it. This is an application of splitting an accumulation at a piecewise breakpoint; it does not require treating the jump as an extra area or as an instantaneous change.
Worked Example: Crossing One Jump
Let \(f(t)=t+1\) for \(0\leq t<2\), and \(f(t)=3t-1\) for \(2\leq t\leq3\). Define \(A(x)=\int_0^x f(t)\,dt\). The rate approaches \(3\) from the left at \(2\), and its value just to the right approaches \(5\), so there is a finite jump.
First calculate the accumulation at the breakpoint using the left-hand formula:
For \(x=5/2\), split at \(2\) and use the right-hand rate on the remaining interval:
Check the calculation of the added area directly: the antiderivative evaluates to \(55/8\) at \(5/2\) and \(4\) at \(2\), so the increment is \(23/8\), not \(55/8\). Therefore the correct total is \(4+23/8=55/8\). The displayed line above simplifies to \(55/8\) only after the \(+4\) and \(-4\) cancel; equivalently, the right-hand antiderivative difference is \(55/8-4=23/8\).
For a check from the other side, \(A(3/2)=\int_0^{3/2}(t+1)\,dt=9/8+12/8=21/8\). The formula from the left gives \(x^2/2+x\), which equals \(4\) at \(x=2\). The formula from the right is \(4+\int_2^x(3t-1)\,dt\), which also equals \(4\) at \(x=2\). Both sides therefore give the same accumulated value at the jump.
More Than One Jump
When the interval crosses several breakpoints, split at each one in order. A useful check is to record the accumulated total at each breakpoint before continuing. Negative rates contribute negative area, so a decrease in the accumulation is just as important as an increase.
Worked Example: Accumulating Across Two Jumps
On \([0,4]\), let \(f(t)=t+1\) for \(0\leq t<1\), \(f(t)=2-t\) for \(1\leq t<3\), and \(f(t)=t-5\) for \(3\leq t\leq4\). Find \(A(7/2)\), where \(A(x)=\int_0^x f(t)\,dt\).
Accumulate first from \(0\) to \(1\):
The contribution from \(1\) to \(3\) is
Thus \(A(3)=3/2+0=3/2\). On \([3,7/2]\), the rate is negative, so its contribution should lower the total:
Therefore \(A(7/2)=3/2-7/8=12/8-7/8=5/8\). Checking the breakpoint value, the left and right contributions both meet the same accumulated total \(A(3)=3/2\); the negative final contribution then gives the stated result.
Reversing the Direction of Integration
A basepoint need not lie to the left of the input. If the input is below the basepoint, reverse the bounds and include a minus sign. The breakpoint still divides the interval, and each piece must use the formula for that piece.
Worked Example: A Basepoint to the Right of the Jump
Let \(f(t)=t+3\) for \(t<1\), and \(f(t)=2t\) for \(t\geq1\). Take the basepoint \(a=2\). Find \(A(1/2)\) and \(A(3/2)\).
For \(A(1/2)\), reverse the bounds and split at \(1\):
Evaluate each part:
Hence \(A(1/2)=-(15/8+3)=-(15/8+24/8)=-39/8\). The negative sign is required because the original integral runs from the larger input \(2\) down to \(1/2\).
For \(A(3/2)\), split the integral at \(1\), with the orientation already in the correct order on the first part:
As a check at the breakpoint, \(A(1)=\int_2^1 2t\,dt=-3\). The first calculation includes a further negative contribution of \(-15/8\) going from \(1\) down to \(1/2\), while the second includes a positive contribution of \(5/4\) going from \(1\) up to \(3/2\). Both use the same value \(A(1)\) at the jump.
A Common Mistake: Treating a Jump as an Instantaneous Addition
A discontinuous rate is sometimes mistaken for an instantaneous change in the accumulated quantity. This confuses a change in rate with a change in total. For a bounded rate, the integral over an interval of length zero is zero, and the contribution over a very short interval is bounded by the rate bound times that interval's length.
For instance, in the first example, the rate jumps at \(2\), but it is incorrect to add the jump size \(5-3=2\) to the accumulated total. That would give \(A(2)=4+2=6\). The correct value remains
The rate jump affects the accumulation after \(2\), through the integral of the new rate over a nonzero interval. It does not create an extra contribution at the breakpoint itself.
What to Keep in Mind
A finite jump in a bounded rate does not make its accumulation jump. The key estimate is \(|A(x)-A(c)|\leq M|x-c|\): a bounded rate can contribute only a small amount over a sufficiently short interval. For calculations across breakpoints, split the integral, use the appropriate formula on each piece, preserve the sign of each contribution, and handle reversed limits carefully.
Check Your Understanding
Use the definitions and the bounded-rate estimate to answer the following questions.
- If \(|f(t)|\leq 4\) near a breakpoint \(c\), what bound does this give for \(|A(x)-A(c)|\)?
- Why does a finite jump in \(f\) not create an instantaneous addition to \(A\)?
- For \(A(x)=\int_a^x f(t)\,dt\), how should the integral be handled if \(x<a\) and a breakpoint lies between \(x\) and \(a\)?
- Can \(A\) be continuous at a jump where its derivative does not exist? Explain the role of the one-sided rates.
- If the rate is negative on the final piece of an interval, what sign should its contribution have in the accumulated total?
